difficult：easy #1
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:

Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

解法1 暴力法

时间复杂度 O($n^2$)

std::vector<int> sum::twoSum(std::vector<int>& nums, int target){
    std::vector<int> result;
    for (int i = 0; i < nums.size(); i++) {
      for (int j = i + 1; j < nums.size(); j++) {
        int sum = nums[i] + nums[j];
        if (sum == target) {
          result.push_back(i);
          result.push_back(j);
        }
      }
    }
    return result;
}

解法2 HashMap

时间复杂度 O($n$)

vector<int> twoSum(vector<int>& nums, int target) {
    std::map<int,int> nums_map;
    std::vector<int> result;
    for (int i = 0; i < nums.size(); i++){
        nums_map[nums[i]] = i;
    }
    for (int i = 0; i < nums.size(); i++){
        int value = target - nums[i];
        if (nums_map.find(value) != nums_map.end() && nums_map[value] != i) {
            result.push_back(i);
            result.push_back(nums_map[value]);
            return result;
        }
    }
}

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